1.链接地址:
http://bailian.openjudge.cn/practice/1979
http://poj.org/problem?id=1979
2.题目:
- 总时间限制:
- 1000ms
- 内存限制:
- 65536kB
- 描述
- There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can't move on red tiles, he can move only on black tiles.
Write a program to count the number of black tiles which he can reach by repeating the moves described above.- 输入
- The input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20.
There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows.
'.' - a black tile
'#' - a red tile
'@' - a man on a black tile(appears exactly once in a data set)
The end of the input is indicated by a line consisting of two zeros.- 输出
- For each data set, your program should output a line which contains the number of tiles he can reach from the initial tile (including itself).
- 样例输入
6 9 ....#. .....# ...... ...... ...... ...... ...... #@...# .#..#. 11 9 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#..@#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### ...@... ###.### ..#.#.. ..#.#.. 0 0- 样例输出
45 59 6 13- 来源
- Japan 2004 Domestic
3.思路:
4.代码:
1 #include <iostream> 2 #include <cstdio> 3 4 using namespace std; 5 6 int f(char **arr,const int i,const int j,int w,int h) 7 { 8 int res = 0; 9 if(arr[i][j] == '.') 10 { 11 res += 1; 12 arr[i][j] = '#'; 13 if(i > 0) res += f(arr,i - 1,j,w,h); 14 if(i < h - 1) res += f(arr,i + 1,j,w,h); 15 if(j > 0) res += f(arr,i,j - 1,w,h); 16 if(j < w - 1) res += f(arr,i,j + 1,w,h); 17 } 18 return res; 19 } 20 21 22 int main() 23 { 24 //freopen("C:\Users\wuzhihui\Desktop\input.txt","r",stdin); 25 26 int i,j; 27 28 int w,h; 29 //char ch; 30 while(cin>>w>>h) 31 { 32 if(w == 0 && h == 0) break; 33 //cin>>ch; 34 35 char **arr = new char*[h]; 36 for(i = 0; i < h; ++i) arr[i] = new char[w]; 37 38 for(i = 0; i < h; ++i) 39 { 40 for(j = 0; j < w; ++j) 41 { 42 cin>>arr[i][j]; 43 } 44 //cin>>ch; 45 } 46 47 for(i = 0; i < h; ++i) 48 { 49 for(j = 0; j < w; ++j) 50 { 51 if(arr[i][j] == '@') 52 { 53 arr[i][j] = '.'; 54 cout << f(arr,i,j,w,h) << endl; 55 break; 56 } 57 } 58 if(j < w) break; 59 } 60 61 for(i = 0; i < h; ++i) delete [] arr[i]; 62 delete [] arr; 63 } 64 65 return 0; 66 }